A gas at p 30 kPa and T 260 K has a mass density of 832102

A gas at p = 30 kPa and T = 260 K has a mass density of 8.32×102 kg/m3 . What is the mean free path of the atoms in the gas? Assume that a typical atomic radius is r0.51010m.

Solution

Here,

Pressure , P = 30 *10^3 Pa

T = 260 K

density , p = 8.32 *10^-2 Kg/m^3

radius, r = 0.5 *10^-10 m

mean free path = RT/(sqrt(2) * pi * (2r)^2 * Na *P)

mean free path = 8.314 * 260/(sqrt(2) * 4 * (0.5 *10^-10)^2 * 6.022 *10^23 * 30 *10^3)

solving

mean free path = 8.461 *10^-6 m

the mean free path is 8.461 *10^-6 m